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Math Help - Arithmetic sum to nth term

  1. #1
    Member Greengoblin's Avatar
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    Arithmetic sum to nth term

    I know that in an arithmetic progression, the value of the nth term is given by:

    u_{n}=a+(n-1)d

    where a is the first term and d is the common difference. This is pretty straightforward. I don't understand how this is then used to obtain the formula for the sum of the first n terms:

    S_{n}=\frac{n}{2} [2a+(n-1)d]

    Can someone please explain where this equation comes from, and how it is derived? I'm guessing it's not a big leap from the first one I gave, but it's big enough for me to to work it out. Cheers!
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  2. #2
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    Quote Originally Posted by Greengoblin View Post
    I know that in an arithmetic progression, the value of the nth term is given by:

    u_{n}=a+(n-1)d

    where a is the first term and d is the common difference. This is pretty straightforward. I don't understand how this is then used to obtain the formula for the sum of the first n terms:

    S_{n}=\frac{n}{2} [2a+(n-1)d]

    Can someone please explain where this equation comes from, and how it is derived? I'm guessing it's not a big leap from the first one I gave, but it's big enough for me to to work it out. Cheers!
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